Pipeline q21 Ca

Question

Consider 2 pipelines A and B where pipeline A having 8 stages of uniform delay of 2 ns.  the pipeline B is having 5 stages with respective stage delays 2 ns, 3 ns, 1 ns, 2 ns ,2 ns. How much time ia saved if 100 instructions are pipelined using A instead of B.

a. 90 ns       b. 98 ns        c. 88 ns       d. 0 ns

Answer 1

Time required i pipeline execution (T_p)={ K + ( N - 1 ) }*  t_p 

N=Number of instructions

K=Number of stages

tp =time period for the clock cycle in pipeline ={ MAX(t_i) from i=1 to k +  Buffer delay }

For A

Time required in pipeline execution (T_pA)={ 8 + (100-1) } *2

=214 ns

For B

Time required i pipeline execution (T_pB)={5+(100-1) } *3 

=312 ns

time saved =312 - 214 = 98 ns